Question

The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is _________________.

Answer 1

Let r be the radius of the sphere at any time t.

Volume of the sphere, V = $\frac{4}{3}\mathrm{\pi }{r}^3$

$V=\frac{4}{3}\mathrm{\pi }{r}^3$

Differentiating both sides with respect to r, we get

$\frac{dV}{dr}=\frac{4}{3}\mathrm{\pi }\times \frac{d}{dr}{r}^3$

$\Rightarrow \frac{dV}{dr}=\frac{4}{3}\mathrm{\pi }\times 3r^2$

$\Rightarrow \frac{dV}{dr}=4\mathrm{\pi }{r}^2$

Surface area of the sphere, S = $4\mathrm{\pi }{r}^2$

$S=4\mathrm{\pi }{r}^2$

Differentiating both sides with respect to r, we get

$\frac{dS}{dr}=4\mathrm{\pi }\times \frac{d}{dr}{r}^2$

$\Rightarrow \frac{dS}{dr}=4\mathrm{\pi }\times 2r$

$\Rightarrow \frac{dS}{dr}=8\mathrm{\pi }r$

Now, rate of change of volume of a sphere with respect to its surface area

$$\begin{gathered} =\frac{dV}{dS} \\ =\frac{{\displaystyle \frac{dV}{dr}}}{{\displaystyle \frac{dS}{dr}}} \\ =\frac{4\mathrm{\pi }{r}^2}{8\mathrm{\pi }r} \\ =\frac{r}{2} \end{gathered}$$

When r = 2 cm, we get

${\left(\frac{dV}{dS}\right)}_{r=2\mathrm{cm}}=\frac{2\mathrm{cm}}{2}$ = 1 cm

Thus, the rate of change of volume of a sphere with respect to its surface area when the radius is 2 cm is 1 cm.


The rate of change of volume of a sphere with respect to its surface area, when the radius is 2 cm, is ____1 cm____.